SOLUTION: Solve the equation on the interval [0,2pi) 2sin^2x=-3sinx+5, in terms of pi and use integers and/or fractions I have tried making the equation equal to 0, but when I try to sim

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation on the interval [0,2pi) 2sin^2x=-3sinx+5, in terms of pi and use integers and/or fractions I have tried making the equation equal to 0, but when I try to sim      Log On


   

Question 886036: Solve the equation on the interval [0,2pi)
2sin^2x=-3sinx+5, in terms of pi and use integers and/or fractions
I have tried making the equation equal to 0, but when I try to simplify, the sin^2x is confusing me. Please explain how this problem works out. Thank you in advance!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equation on the interval [0,2pi)
2sin^2x=-3sinx+5, in terms of pi and use integers and/or fractions
***
2sin%5E2%28x%29=-3sin%28x%29%2B5
2sin%5E2%28x%29%2B3sin%28x%29-5=0
%282sin%28x%29%2B5%29%28sin%28x%29-1%29=0
2sin(x)+5=0
sin(x)=-5/2(reject,(-1 < sin(x) < 1))
or
sin(x)-1=0
sin(x)=1
x=π/2